Now we are ready to begin our division problem. Just like in multiplication, we will work through the dividend 1 digit at a time using 1 divisor digit at a time. Again we will make full use of our 9×9 multiplication facts to determine each subtraction from the dividend. So we start our problem by noticing that our divisor, 8, is larger than the first digit of the dividend 2. So we can not divide 8 into 2. Therefore we will move 1 more digit right on the dividend and consider the first two digits of the dividend 28. Now we know from our multiplication facts that 8×3 = 24 and 8×4 = 32. So the largest number of times we can take 8 from 28 is 3. Therefore we will place our first quotient digit, 3, on the second rod left from the quotient unit rod.

Here we should make clear how we are maintaining correct place value for the quotient as we work through the dividend. Our dividend has 3 digits so the largest our quotient can be is 3 digits. However, in this problem since we could not divide 8 into the 2, the first dividend digit, we moved 1 more digit to the right on the dividend. Therefore we also have to move 1 digit to the right on the quotient. This is why we placed the first quotient digit 3 on the second rod of the quotient, not the third rod from the quotient unit rod. So we know already our answer will only have 2 digits.